Question

Let A = R – {5} and B = R – {1}. Consider the function f : A → B, defined by `f(x) = (x-3)/(x-5)`. Show that f is one-one and onto.

Answer 1

Given, A = R − {5}, B = R − {1}

f: A → B

`f(x) = (x-3)/(x-5)`

for f is one-one

Let x₁, x₂ ∈ R

f(x₁) = f(x₂)

`(x_1-3)/(x_1-5) = (x_2-3)/(x_2-5)`

(x₁ − 3) (x₂ − 5) = (x₂ − 3) (x₁ − 5)

x₁x₂ − 5x₁ − 3x₂ + 15 = x₁x₂ − 5x₂ − 3x₁ + 15

−5x₁ + 3x₁ = −5x₂ + 3x₂

−2x₁ = −2x₂

∴ x₁ = x₂

f is one-one. 
For if is onto

Let y be any element of R

y = f(x)

⇒ `y = (x-3)/(x-5)`

⇒ y(x − 5) = x − 3

⇒ yx − 5y = x − 3

⇒ yx − x = 5y − 3

⇒ `x = (5y-3)/(y-1)`

`f(x) = (x-3)/(x-5)`

`f((5y-3)/(y-1))= ((5y-3)/(y-1) - 3)/((5y-3)/(y-1) - 5)`

⇒ `f((5y-3)/(y-1)) = ((5y-3-3y+3)/(y-1))/((5y-3-5y+5)/(y-1))`

`f ((5y-3)/(y-1)) = (2y)/2 = y`

f is onto